Let $A_1,\ldots,A_n$ be $n$ events in a discrete probability space. Can someone give me an example such that $$P(A_1 \cap \cdots \cap A_n)=P(A_1)\cdot \ldots \cdot P(A_n)$$ holds, but such that there is a subset $A_{i_1},\ldots,A_{i_k}$of the $A_1,\ldots,A_n$, such that $$P(A_{i_1}\cap\ldots \cap A_{i_k})\neq P(A_{i_1})\cdot\ldots \cdot P(A_{i_k}) \ \ ?$$

If that weren't such an example that would mean that the usual definition of mutual independence - as here, for example - is too strong, since the independence of the intersection of all events already implies the independence of all partial intersections of events. In that case I would like a proof of this implication.

2 Answers
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The subset $A_{i_1}, \ldots, A_{i_k}$ can be pretty far from independent — they can be the same event! All you need is for the remaining events to be such that the net intersection is small enough.

For example, consider the probability space of three fair coin tosses and let $A_1=A_2=\{\text{first toss is heads}\}$. For your condition to hold, we can just take an $A_3$ such that $P(A_3)=1/2$ but $P(A_1\cap A_3)=1/8$; in particular, $A_3=\{\text{at least two tosses are tails}\}$ will do. Then
$$P(A_1\cap A_2\cap A_3) = P(\{\text{heads-tails-tails}\}) = \frac18 = P(A_1)\cdot P(A_2)\cdot P(A_3),$$
but $A_1$ and $A_2$ are obviously not independent.